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3 years ago

Which of the following is an example of a wave?

Physics

2 answers:

Pavlova-9 [17]3 years ago

6 0

Answer:

A plucked guitar string

Explanation:

When you pluck a guitar string it causes a sound wave which is also an example of a longitudinal wave.

qaws [65]3 years ago

3 0

The answer is A, it’s a sound wave

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What is the relationship between potential and kinetic energy in chemical reactions?

astraxan [27]

Answer:

2 and 3

Explanation:

because that's how kinetic and potential energy work

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3 years ago

Near the equator, rising air is associated with a pressure zone known as the _____.

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4 years ago

Read 2 more answers

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (6)

Where:

$a=-4.9 m/s^{2}$

$b=48.1 m/s$

$c=1.5 m$

Substituting the known values:

$t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)}$ (7)

Solving (7) we find the positive result is:

$t=9.847 s$ (8)

Substituting this value in (1):

$x=(48.1 m/s)cos(0\°) (9.847 s)$ (9)

$x=473.640 m$ This is the horizontal distance the cannonball traveled before it landed on the ground.

3 0

3 years ago

When ionic bonds form compound structures in a solid state, they arrange in latticed, crystalline patterns.

Phantasy [73]

Answer:

Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound.

5 0

3 years ago

What are the magnitude Δ v and angle θ of the change in velocity if the initial velocity is 50.0 m/s south and the final velocit

I am Lyosha [343]

Answer:

∆v = 61.0 m/s

θ = 305°

Explanation:

Given;

initial velocity u= 50.0 m/s angle 270°

Final velocity v = 35.0 m/s angle 180°

∆v = v - u ........1

Resolving into x and y axis;

For y axis

∆vy = vsin180 - usin270

∆vy = 0 - (-50) = 50 m/s

x axis;

∆vx = vcos180 - ucos270

∆vx = -35 - 0= -35m/s

Magnitude of ∆v;

∆v = √(-35^2 + 50^2)

∆v = 61.0 m/s

Direction;

Tanθ = ∆vy/∆vx

θ = taninverse (50/-35)

θ = -55°

θ = 360-55 = 305°

5 0

4 years ago